InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is a digital circuit? |
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Answer» An electronic circuit that processes only digital signals is called a digital circuit. There are only two voltage states present at any point within a digital circuit. These voltage states are either high or low. The branch of electronics which deals with digital circuits is called digital electronics. [Note : Digital circuits can store and process bits of information needed to make complex logical decisions. Digital electronics incorporate logical decision-making processes into a circuit.] |
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| 2. |
How many rows are there in the truth table of a 3-input gate? |
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Answer» Each of the 3 inputs can take 2 values (0 and 1). Hence, the number of rows in the truth table of a 3-input gate = 2 × 2 × 2 = 22 = 8. |
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| 3. |
What are the three ways of representing a logic gate? |
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Answer» A logic gate can be represented by its logic symbol, Boolean expression and the truth table. |
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| 4. |
White Light LEDs or White LED Lamps: Shuji Nakamura, a Japanese – born American electronic engineer invented the blue LED. He was awarded the Nobel prize for physics for 2014. He was also awarded the global energy prize in the year 2015. His invention of blue LED made the fabrication of white LED possible.LED lamps, bulbs, street lighting are becoming very popular these days because of the very high efficiency of LEDs in terms of light output per unit input power(in milliWatts), as compared to the incandescent bulbs. So for general purpose lightings, white light is preferred.Commercially available white LEDs are normally manufactured by using the technique of wavelength conversion. It is a process which partly or completely converts the radiation of a LED into white light. There are many ways of wavelength conversion. One of these methods uses blue LED and yellow phosphor. In this method of wavelength conversion, a LED which emits blue colour is used to excite a yellow colour phosphor. This results in the emission of yellow and blue light and this mixture of blue and yellow light gives the appearance of white light. This method is the least expensive method for producing white light. |
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Answer» The all important blue LEDs The development of LEDs has made more efficient light sources possible. Creating white light that can be used for lighting requires a combination of red, green and blue light. Blue LEDs proved to be much more difficult to create than red and green LEDs. During the 1980s and 1990s Isamu Akasaki, Hiroshi Amano, and Shuji Nakamura successfully used the difficult-to-handle semiconductor gallium nitride to create efficient blue LEDs. Isamu Akasaki is known for invent ing the bright gallium nitride (Can) pn-junction blue LED in 1989 and subsequently the high-brightness Can blue LED. Using blue LEDs, highly efficient white light sources. became possible by converting part of the blue light emitted from an LED to yellow using a phosphor. To the human eye, the combination of blue and yellow light is perceived as white. A white LED can be created by embedding phosphors in the plastic cap which surrounds a blue LED. Higher quality white light can also be created by mixing blue light with other colors as well, including red and green. Isamu Akasaki, together with Shuji Nakamura and Hiroshi Amano, received the 2014 Nobel Prize in Physics for the invention of efficient blue light-emitting diodes which has enabled bright and energy saving white light sources. |
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| 5. |
The logic gate which produces LOW output when any one of the input is HIGH and produces HIGH output only when all of its inputs are LOW is called(A) an AND gate (B) an OR gate (C) a NOR gate (D) a NAND gate. |
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Answer» Correct option is (C) a NOR gate |
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| 6. |
What would happen if both junctions of a BJT are forward biased or reverse biased? |
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Answer» A BJT has four regimes of operation, depending on the four combinations of the applied biases (voltage polarities) to the emitter-base junction and the collector-base junction, as shown in the following table; ‘F’ and ‘R’ indicate forward bias and reverse bias, respectively. |
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| 7. |
The Boolean expression for an Exclusive OR gate is (A) A + B (B) A ⊕ B(C) \(\overline{A+B}\)(D) A ∙ B |
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Answer» Correct option is (B) A ⊕ B |
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| 8. |
Which logic gate corresponds to the truth tableABY001010100110(A) AND (B) NOR (C) OR (D) NAND |
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Answer» Correct option is (B) NOR |
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| 9. |
The lightly doped, thin base region sandwiched between the heavily doped emitter region and the intermediate doped collector region plays a crucial role in the transistor action. |
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Answer» If the two junctions in a BJT are physically close compared with the minority carrier diffusion length (i.e., the distance within which recombination will take place), the careers injected from the emitter can diffuse through the base to reach the base-collector junction. The narrow width of the base is thus crucial for transistor action. |
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| 10. |
An LED emits visible light when its(A) junction is reverse biased(B) depletion region widens(C) holes and electrons recombine(D) junction becomes hot. |
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Answer» Correct option is (C) holes and electrons recombine |
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| 11. |
In a BJT, the largest current flow occurs (A) in the emitter (B) in the collector (C) in the base (D) through CB junction. |
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Answer» Correct option is (A) in the emitter |
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| 12. |
Zener effect occurs only if the diode is heavily doped, because when the depletion layer is thin, breakdown occurs at low reverse voltage and the field strength will be approximately 3 × 107 V/m. It causes an increase in the flow of free carriers and increase in the reverse current. |
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Answer» Zener breakdown occurs only in heavily doped pn-junctions (doping concentrations for both p- and n-regions greater than 1018 cm3) and can take place only if the electric field in the depletion region of the reverse-biased junction is very high. It is found that the critical field at which tunneling becomes probable, i.e., at which Zener breakdown commences, is approximately 106 V/cm. |
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| 13. |
A series resistance is connected in the Zener diode circuit to (A) properly reverse bias the Zener (B) protect the Zener (C) properly forward bias the Zener(D) protect the load resistance. |
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Answer» Correct option is (A) properly reverse bias the Zener |
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| 14. |
The stepped down output of a transformer, with turns ratio 5 : 1, is fed to an ideal diode D and load resistance RL. The peak load voltage is(A) 22 V (B) 31 V (C) 44 V (D) 62 V |
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Answer» Correct option is (D) 62 V |
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| 15. |
When the intensity of light incident on a photo diode increases, how is the reverse current affected? |
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Answer» The photocurrent increases linearly with increasing illuminance, limited by the power dissipation of the photodiode. |
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| 16. |
Why is a resistance connected in series with a Zener diode when used in a circuit? |
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Answer» The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current IZ can rapidly increase at constant VZ . To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, IZM . Hence, a current-limiting resistor RS is connected in series with the diode. IZ and the power dissipated in the Zener diode will be large for IL = 0 (no-load condition) or when IL is less than the rated maximum (when RS is small and RL is large). The currentlimiting resistor RS is so chosen that the Zener current does not exceed the rated maximum reverse current, IZM when there is no load or when the load is very high. The rated maximum power of a Zener diode is PZM = IZM = VZ At n-load condition, the current through R is IZM = I and the voltage drop across it is V – VZr , where V is the unregulated source voltage. The diode current will be maximum when V is maximum at Vmax and I = IZM . Then, the minimum value of the series resistance should be RS, min = \(\cfrac{V_{max}-V_Z}{I_{ZM}}\) |
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| 17. |
What is the difference between a photo diode and a solar cell? |
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Answer» Both are semiconductor photovoltaic devices. A photodiode is a reverse-biased pn-junction diode while a solar cell is an unbiased pn junction diode. Photodiodes, however, are optimized for light detection while solar cells are optimized for energy conversion efficiency. |
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| 18. |
Explain the use of a resistor in series with a Zener diode. |
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Answer» The I-V characteristics in the breakdown region of a Zener diode is almost vertical. That is, the current IZ can rapidly increase at constant VZ. To prevent damage due to excessive heating, the Zener current should not exceed the rated maximum current, IZM. Hence, a current-limiting resistor Rs is connected in series with the diode IZ and the power dissipated in the Zener diode will be large for IL = 0 (no-load condition) or when IL is less than the rated maximum (when Rs is small and RL is large). The currentlimiting resistor RS is so chosen that the Zener current does not exceed the rated maximum reverse current, IZM when there is no load or when the load is very high. The rated maximum power of a Zener diode is PZM = IZM = VZ At no-load condition, the current through Rs Z = IZM and the voltage drop across it is V – VZ, where V is the unregulated source voltage. The diode current will be maximum when V is maxi-mum at Vmax and I = IZM. Then, the minimum value of the series resistance should be Rs, min = \(\cfrac{V_{max}−V_Z}{I_{ZM}}\) A Zener diode is operated in the breakdown region. There is a minimum Zener current, IZ (min), that places the desired operating point in the breakdown region. There is a maximum Zener current, IZM, at which the power dissipation drives the junction temperature to the maximum allowed. Beyond that current the diode can be damaged. Hence, the supply voltage must be greater than Vz and the current-limiting resistor must limit the diode current to less than the rated maximum, IZM. |
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| 19. |
The voltage across a Zener diode does not remain strictly constant with the changes in the Zener current. This is due to RZ, the Zener impedance, or the internal resistance of the Zener diode. RZ acts like a small resistance in series with the Zener. Changes in IZ cause small changes in VZ . |
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Answer» The I-V characteristics of a Zener diode in the breakdown region is not strictly vertical. Its slope is 1/RZ , where RZ is the Zener impedance. |
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| 20. |
State any two applications of a Zener diode. |
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Answer» Applications of a Zener diode : 1. Voltage regulator 2. Fixed reference voltage in biasing transistors 3. Peak clipper in a wave shaping circuit 4. Meter protection from voltage fluctuations. |
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| 21. |
The photocurrent in a photodiode is a few (A) nanoamperes (B) microamperes (C) milliamperes (D) amperes. |
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Answer» (B) microamperes |
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| 22. |
A Zener diode has a Zener voltage of 2.4 V and a 500 mW power rating. What should be the maxi-mum current through the diode if you design conservatively with a safety factor of 2? |
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Answer» Data : Vz = 2.4 V, PZM = 500 mW A conservative design includes a safety factor to keep the power dissipation well below the rated maximum power. Thus, with a safety factor of 2, the operating power of the Zener diode is PZ = \(\cfrac12\) PZM = \(\cfrac{500}2\) = 250 m W Therefore, the Zener current, IZ = \(\cfrac{p_z}{V_z}\) = \(\cfrac{250mW}{2.4\,V}\) = 104.2 mA |
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| 23. |
In operation, a photodiode is (A) unbiased (B) always forward-biased (C) always reverse-biased (D) either forward-or reverse-biased. |
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Answer» (C) always reverse-biased |
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| 24. |
Explain Zener breakdown. |
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Answer» In a reverse-biased pn-junction, the depletion region is wider and the potential barrier is higher over equilibrium values. The electric field in the depletion region is from the n- to the p-region. When a sufficiently large reverse voltage is applied to a pn-junction, the junction breaks down and conducts a very large current. Of the two important breakdown mechanisms, Zener breakdown takes place in heavily doped diodes. Usually, the energy that an electron can gain from even a strong field is very small. However, the depletion region is very narrow in a heavily doped diode. Because of this, the electric field across the depletion region is intense enough to break the covalent bonds between neighbouring silicon atoms and pull electrons out of their orbits. This results in conduction electrons and holes. In the energy band diagram representation, this corresponds to the transition of an electron from the valence band to the conduction band and become available for conduction. The current increases with increase in applied voltage, but without further increase in voltage across the diode. This process, in which an electron of energy less than the barrier height penetrates through the energy bandgap, is called tunneling (a quantum mechanical effect). The creation of electrons in the conduction band and holes in the valence band by tunneling effect in a reverse- biased pn-junction diode is called the Zener effect. [Notes : (1) Tunneling occurs only if the electric field is very high. The typical field for silicon and gallium arsenide is > 106 V / cm. To achieve such a high field, the doping concentrations for both p- and w-regions must be quite high (>1018 cm-3). Zener breakdown or Zener effect is named in honour of Clarence M. Zener (1905-93), US physicist, who explained the breakdown mechanism. (3) Avalanche breakdown occurs in diodes with a doping concentration of ≅ 1017 cm-3 or less. The carriers gain enough kinetic energy to generate electron-hole pairs by the avalanche process when the value of reverse | V | becomes large. An electron in the conduction band can gain kinetic energy before it collides with a valence electron. The high-energy electron in the conduction band can transfer some of its kinetic energy to the valence electron to make an upward transition to the conduction band. An electron-hole pair is generated. All such electrons and holes accelerate in the high field of the depletion region and, in turn, generate other electronhole pairs in a like manner. This process is called the avalanche process.] |
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| 25. |
The current through the Zener diode in the following circuit is(A) 5 mA(B) 10 mA (C) 15 mA (D) 30 mA. |
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Answer» Correct option is (A) 5 mA |
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| 26. |
Avalanche breakdown in a Zener diode takes place due to (A) thermal energy (B) light energy (C) magnetic field(D) accelerated minority charge carriers. |
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Answer» (D) accelerated minority charge carriers. |
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| 27. |
State any two advantages of a full-wave rectifier. |
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Answer» 1. A full-wave rectifier rectifies both halves of each cycle of the ac input. 2. Efficiency of a full-wave rectifier is twice that of a half-wave rectifier. 3. The ripple in a full-wave rectifier is less than that in a half-wave rectifier. Ripple factors for a full-wave and halfwave rectifiers are respectively, 0.482 and 1.21. |
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| 28. |
The maximum efficiency of a full wave rectifier is 81.2% and the maximum efficiency of a half wave rectifier is 40.6%. It is observed that the maximum efficiency of a full wave rectifier is twice that of half wave rectifier. |
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Answer» The ratio of dc power obtained at the output to the applied input ac power is known as rectifier efficiency. A half-wave rectifier can convert maximum 40.6% of ac power into dc power, and the remaining power of 59.4% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit. Hence, a half wave rectifier efficiency is 40.6%. The maximum efficiency of a full-wave rectifier is 81.2%, i.e., twice that of a half-wave rectifier. |
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| 29. |
A full-wave rectifier uses a grounded centre tap on the secondary winding of a transformer and a 60 Hz source voltage across the primary winding. The output frequency of the fullwave rectifier is(A) 30 Hz (B) 60 Hz (C) 120 Hz (D) 240 Hz. |
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Answer» Correct option is (C) 120 Hz |
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| 30. |
Distinguish between a half-wave rectifier and full-wave rectifier. |
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Answer»
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| 31. |
In a common-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current. |
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Answer» DATA : IE = 10 mA, IC = 9.8 mA IE = IB + IC Therefore, the base current, IB = IE – IC = 10 – 9.8 = 0.2 mA |
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| 32. |
A transistor acts as a ‘closed switch’ when it is in (A) the cutoff region (B) the active region (C) the breakdown region (D) the saturation region. |
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Answer» (D) the saturation region. |
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| 33. |
What is a light-emitting diode (LED) ? |
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Answer» A light-emitting diode (LED) is a forwardbiased pn-junction diode formed from compound semiconductor materials such as gallium arsenide (GaAs) in which light emission can take place from direct radiative recombination of excess electron-hole pairs. A photon is emitted when an electron in the conduction band recombines with a hole in the valence band. In infrared emitting LEDs, the encapsulating plastic lens may be impregnated or coated with phosphorus. Then, phosphorescence of the phos-phorus gives off visible light. [Note: In an ordinary pn-junction diode, energy released in electron-hole recombination process is absorbed in the crystal structure as heat.] |
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| 34. |
On which factors does the wavelength of light emitted by a LED depend? |
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Answer» The intensity of the emitted light is directly propor-tional to the recombination rate and hence to the diode forward current. The colour of the light emitted by an LED depends on the compound semiconductor material used and its composition (and doping levels) as given below : Table: Typical semiconductor materials and emitted colours of LEDs
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| 35. |
How is a Zener diode different than an ordinary diode? |
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Answer» A Zener diode is heavily doped-the doping con-centrations for both p- and n-regions is greater than 1018 cm-3 while those of an ordinary diode are voltage (PIV) of an ordinary diode is higher than a Zener diode and the breakdown occurs by impact ionization (avalanche process). Their I-V characteristics are otherwise similar |
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| 36. |
Why do we need filters in a power supply? |
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Answer» A rectifier-half-wave or full-wave – outputs a pul-sating dc which is not directly usable in most electronic circuits. These circuits require something closer to pure dc as produced by batteries. Unlike pure dc waveform of a battery, a rectifier output has an ac ripple riding on a dc waveform. The circuit used in a dc power supply to remove the ripple is called a filter. A filter circuit can produce a very smooth waveform that approximates the waveform produced by a battery. The most common technique used for filtering is a capacitor connected across the output of a rectifier. |
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| 37. |
Make a list of at least 5 duties performed by an automobile sales person. |
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Answer» 1. Approaches, greets and offers assistance or direction to any customer who enters the dealership. 2. Assists customers in selecting a vehicle by asking questions and listening carefully to their responses. 3. Explains fully the product performance, application and benefits. 4. Describes all optional equipment available for customer purchase. 5. Utilizes dealership sales control and follow-up system. 6. Exhibits high level of commitment to customer satisfaction. 7. Introduces customers to service department personnel. 8. Schedules first service appointment. 9. Follows up on all post-delivery items, tag/title work, “we-owes”, and special requests to be sure that all customer expectations are met. 10. Maintains an owner follow-up system that encourages repeat and referral business and contributes to customer satisfaction. 11. Maintains a prospect development system. 12. Reviews and analyzes actions at the end of each day, week, month, and year to determine how to better utilize time and plans more effectively. 13. Attends sales meetings. 14. Maintains a well-groomed and professional appearance. |
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| 38. |
The diagram shows a CE circuit using a silicon transistor. Calculate the (a) base current (b) collector current. [VBB = 2V, VCC = 10V, RB = 100 kΩ, RL = 1 kΩ, [βdc = 200] |
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Answer» Data : VBB = 2 V, VCC = 10 V, RB = 100 kΩ, RL = 1 kΩ, βdc = 200. Since it is a silicon transistor, the emitter-base barrier potential, V = 0.7 V. The voltage across the base resistor is VBB – VBE = 2 – 0.7 = 1.3 V Therefore, the base current, IB = \(\cfrac{V_{BB}−V_{BE}}{R_B}\) = \(\cfrac{1.3}{10^5}\) = 1.3 x 10-5 = 13 μA The collector current, IC = βIB = 200 × 1.3 × 10-5 = 2.6 × 10-3 A = 2.6mA |
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| 39. |
What is meant by an analog signal and an analog electronic circuit? |
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Answer» An analog signal consists of a continuously varying voltage or current. An analog electronic circuit takes an analog signal as input and outputs a signal that varies continuously according to the input signal. |
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| 40. |
State any two applications of photodiodes. |
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Answer» Applications of photodiodes : 1. A reverse-biased photodiode conducts only when illuminated, assuming that the dark current is essentially zero. Due to its quick response to radi-ation and high operational speed, photodiodes are used in high-speed counting or switching applications. 2. Extensively used in an fibreoptic communication system. 3. As photosensors/photodetectors for detection of UV radiations and accurate measurement of illumination. Avalanche photodiodes have increased responsivity and can be used as photomultipliers, especially for low illumination. 4. In burglar alarm systems as normally closed switch until exposure to radiation is interrupted. When interrupted, the reverse current drops to the dark current level and sounds the alarm. 5. In an optocoupler, a photodiode is combined with a light-emitting diode to couple an input signal to the output circuit. The key advantage of an optocoupler is the electrical isolation between the input and output circuits, especially in high voltage applications. With an optocoupler, the only contact between the input and the output is a beam of light. |
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| 41. |
What is a photodiode? |
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Answer» A photodiode is a special purpose reverse biased pn-junction diode that generates charge carriers in response to photons and high energy particles, and passes a photocurrent in the external circuit proportional to the intensity of the incident radiation. The term photodiode usually means a sensor that accurately detects changes in light level. Hence, it is sometimes called a photodetector or photosensor which operates as a photoelectric converter. |
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| 42. |
Name two types of solar energy devices. |
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Answer» Two major types of devices converting solar energy in usable form are 1. photothermal devices, which convert the solar energy into heat energy 2. photovoltaic devices, which convert solar energy into electrical energy. |
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| 43. |
State the principle and uses of a solar Cell. |
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Answer» A solar cell is an unbiased pn junction that converts the energy of sunlight directly into electricity with a high conversion efficiency Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pn junction as a result of irradiation. |
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| 44. |
State the material selection criteria for solar cells. |
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Answer» Criteria for materials to be used in solar cells : 1. Band gap energy must be between 1 eV and 1.8 eV. (The best band gap of a solar cell is in the region of 1.5 eV.) 2. It must have high optical absorption. 3. It must have high electrical conductivity. 4. The raw material must be available in abundance and the cost of the material must be low. |
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| 45. |
When the load resistance across a solar cell is zero, the current in the external circuit passed by the solar cell is called (A) the open-circuit current (B) the reverse saturation current (C) the short-circuit current (D) the photocurrent. |
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Answer» (C) the short-circuit current |
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| 46. |
What is a solar cell ? State the principle of its working. |
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Answer» A solar cell is an unbiased pn-junction that converts the energy of sunlight directly into electricity with a high conversion efficiency. Principle : A solar cell works on the photovoltaic effect in which an emf is produced between the two layers of a pnjunction as a result of irradiation. |
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| 47. |
Name the common materials for solar cells. |
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Answer» Optimized band gap for solar cells is close to 1.5 eV. Some of the common materials for solar cells are 1. silicon (Si), EG = 1.12 eV – currently the most popular material but has low absorption coefficient and high temperature dependence, 2. gallium arsenide (GaAs), EG = 1.42 eV - by far the most widely used, especially for high end applications like satellites. Its absorption coefficient is about ten times better than silicon and doesn’t have the same temperature dependence. 3. copper-indium diselenide (CIS), EG = 1.01 eV – has the highest optical absorption, but gallium is introduced in the lattice to raise the band gap energy closer to the solar ideal. This resulted in the popular copper-indium-gallium diselenide (Culn- GaSe2 or CIGS) material for photovoltaic cell. By variation of Ga fraction, a band gap of around 1.48 eV has been achieved. 4. cadmium telluride (CdTe), EG = 1.44 eVmade from the II-VI group elements. [Notes : (1) By far the most widely used III-V solar cell is gallium arsenide (GaAs). Other IIIV semiconductors-indium phosphide (InP), gallium antimonide (GaSb), aluminium gallium arsenide (AlGaAs), indium gallium phosphide (InGaP), and indium gallium arsenide (InGaAs)-exchange group III elements to make different band gap energies. III-V semiconductors offer a great host of advantages over silicon as a material for photovoltaics. However, the biggest drawback is cost. (2) The theoretical limit on the thermodynamic efficiency of single-junction solar cells is ~ 30%. Hence, today’s most efficient technology for the generation of electricity from solar radiation is the use of multi-junction solar cells made of III-V compound semiconductors. Efficiencies up to 39% have al-ready been reported under concentrated sunlight. These solar cells have initially been developed for powering satellites in space and are now starting to explore the terrestrial energy market through the use of photovoltaic concentrator systems. A triple-junction solar cell, Ga0.35 In0.65 P/Ga0.83 In0.17 As/Ge, has been demonstrated by a conversion efficiency of 41.1% at 454 kW/m2] |
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| 48. |
Name any four common special-purpose diodes. |
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Answer» Special-purpose diodes : 1. Zener diode 2. light emitting diode (LED) 3. photodiode 4. solar cell. |
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| 49. |
What is a unregulated power supply ? |
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Answer» A dc power supply whose output changes when a load is connected across it is called unregulated power supply. |
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| 50. |
What is a regulated power supply? |
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Answer» A dc power supply whose preset output voltage remains constant irrespective of variations in the line voltage or load current is called a regulated power supply. |
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